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∵△AGF≌△EBD {等量减等量差相等AG=EB;同位角∠A=∠BED,∠AGF=∠B;兩角夾一邊}, 故GF=BD{對應邊相等}; ∴DF‖AB{一組對邊平行且相等GF=‖BD,GFDB為平行四邊形}.
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